pandas 在python中在下划线处拆分并存储第一个值

splitting at underscore in python and storing the first value

提问by Ssank

I have a pandas data frame like df with a column construct_name

我有一个像 df 这样的 Pandas 数据框，带有一个列construct_name

construct_name
aaaa_t1_2    
cccc_t4_10
bbbb_g3_3

and so on. I want to first split all the names at the underscore and store the first element (aaaa,cccc, etc.) as another column name.

等等。我想首先在下划线处拆分所有名称，并将第一个元素（aaaa、cccc 等）存储为另一个列名称。

Expected output

预期输出

construct_name  name
aaaa_t1_2       aaaa
cccc_t4_10      bbbb

and so on.

等等。

I tried the following df['construct_name'].map(lambda row:row.split("_"))and it gives me a list like

我尝试了以下操作 df['construct_name'].map(lambda row:row.split("_"))，它给了我一个列表

[aaaa,t1,2]
[cccc,t4,10]

and so on

等等

But when I do

但是当我做

df['construct_name'].map(lambda row:row.split("_"))[0]to get the first element of the list I get an error. Can you suggest a fix. Thanks

df['construct_name'].map(lambda row:row.split("_"))[0]要获取列表的第一个元素，我收到错误消息。你能建议修复吗。谢谢

回答by EdChum

Just use the vectorised strmethod splitand use integer indexing on the list to get the first element:

只需使用矢量化str方法split并在列表上使用整数索引来获取第一个元素：

In [228]:

df['first'] = df['construct_name'].str.split('_').str[0]
df
Out[228]:
  construct_name first
0      aaaa_t1_2  aaaa
1     cccc_t4_10  cccc
2      bbbb_g3_3  bbbb

回答by fixxxer

After you do the split, you should get the first element (using [0]). And not after the map.:

完成后split，您应该获得第一个元素（使用 [0]）。而不是在map. 之后：

In [608]: temp['name'] = temp['construct_name'].map(lambda v: v.split('_')[0])

In [609]: temp
Out[609]: 
  construct_name  name
0      aaaa_t1_2  aaaa
1     cccc_t4_10  cccc
2      bbbb_g3_3  bbbb

回答by BioGeek

splittake an optional argument maxsplit:

split采用可选参数maxsplit：

>>> construct_name = 'aaaa_t1_2'
>>> name, rest = construct_name.split('_', 1)
>>> name
'aaaa'